Borwein-Garvanによる恒等式

$$\begin{array}{rl}\sum _{n\in \mathbb{Z}}(\frac{a{q}^n}{(1-a{q}^n{)}^2}-\frac{b{q}^n}{(1-b{q}^n{)}^2})& =a\frac{(ab,q/ab,b/a,aq/b;q{)}_{\mathrm{\infty }}(q;q{)}_{\mathrm{\infty }}^4}{(a,q/a,b,q/b;q{)}_{\mathrm{\infty }}^2}\end{array}$$
において, $q\mapsto q^3$として$a=\omega :=e^{\frac{2\pi i}{3}},b:=q$として,
$$\begin{array}{rl}(a,a\omega ,a{\omega }^2;q{)}_{\mathrm{\infty }}& =(a^3;q^3{)}_{\mathrm{\infty }}\\ (a,aq,a{q}^2;q^3)& =(a;q{)}_{\mathrm{\infty }}\end{array}$$
などを用いて整理すると,
$$\begin{array}{rl}\sum _{n\in \mathbb{Z}}(\frac{\omega q^{3n}}{(1-\omega q^{3n}{)}^2}-\frac{q^{3n+1}}{(1-q^{3n+1}{)}^2})& =\omega \frac{(\omega q,{\omega }^2{q}^2,{\omega }^{2}q,\omega q^2,q^3,q^3,q^3,q^3;q^3{)}_{\mathrm{\infty }}}{(\omega ,{\omega }^2{q}^3,q,q^2;q^3{)}_{\mathrm{\infty }}^2}\\ & =\frac{\omega }{(1-\omega {)}^2}\frac{(q^3,q^6;q^9{)}_{\mathrm{\infty }}(q^3;q^3{)}_{\mathrm{\infty }}^4}{(\omega q^3,{\omega }^2{q}^3{)}_{\mathrm{\infty }}^2(q,q^2;q^3{)}_{\mathrm{\infty }}^3}\\ & =-\frac{1}{3}\frac{(q^3;q^3{)}_{\mathrm{\infty }}^{10}}{(q^9;q^9{)}_{\mathrm{\infty }}^3(q;q{)}_{\mathrm{\infty }}^3}\end{array}$$
となる. 一方, 左辺は
$$\begin{array}{rl}& \sum _{n\in \mathbb{Z}}(\frac{\omega q^{3n}}{(1-\omega q^{3n}{)}^2}-\frac{q^{3n+1}}{(1-q^{3n+1}{)}^2})\\ & =\frac{\omega }{(1-\omega {)}^2}+\sum _{0\le n}(\frac{\omega q^{3n+3}}{(1-\omega q^{3n+3}{)}^2}+\frac{{\omega }^2{q}^{3n+3}}{(1-{\omega }^2{q}^{3n+3}{)}^2}-\frac{q^{3n+1}}{(1-q^{3n+1}{)}^2}-\frac{q^{3n+2}}{1-q^{3n+3}})\\ & =-\frac{1}{3}+\sum _{0\le n}(\frac{q^{3n+3}}{(1-q^{3n+3}{)}^2}+\frac{\omega q^{3n+3}}{(1-\omega q^{3n+3}{)}^2}+\frac{{\omega }^2{q}^{3n+3}}{(1-{\omega }^2{q}^{3n+3}{)}^2}-\frac{q^{n+1}}{1-q^{n+1}})\\ & =-\frac{1}{3}+\sum _{0\le n}(\frac{9q^{9n+9}}{(1-q^{9n+9}{)}^2}-\frac{q^{n+1}}{1-q^{n+1}})\\ & =-\frac{1}{3}+\sum _{0\le n,0<m}(9m{q}^{(9n+9)m}-m{q}^{(n+1)m})\\ & =-\frac{1}{3}+\sum _{0<m}(\frac{9m{q}^{9m}}{1-q^{9m}}-\frac{m{q}^m}{1-q^m})\end{array}$$
と計算できるので, 両辺に$-3$を掛けて, 定理を得る.